If $a_n \to z$, then does $\frac{{n \choose 1} a_1 + {n \choose 2} a_2 \dots {n \choose n} a_n}{2^n} \to z$?

Something far more general is true. Define the doubly half-infinite matrix $ c_{mn}$, $$ c_{mn} = \frac1{2^m}\binom{m}{n}\mathbb 1_{n\le m} $$ then you're asking if $$ t_m := \sum_{n=0}^\infty c_{mn} a_n \to z?$$ According to Hardy's Divergent Series, Theorem 2 (page 43) (apparently due to Toeplitz and Schur):

Theorem 2. (Paraphrased) For any such infinite matrix $(c_{mn})_{m,n\ge 0}$, the statement $$\text{“$a_n \to z$ implies $t_m \to z$''}$$ is true iff the following conditions for $c_{mn}$ are satisfied:

1. $\sum_{n=0}^\infty |c_{mn} | < H $ where $H<\infty$ is uniform in $m$,
2. for each $n$, $c_{mn} \xrightarrow[m\to\infty]{} 0$,
3. $\sum_{n=0}^\infty c_{mn} \xrightarrow[m\to\infty]{} 1$.

For your specific $c_{mn}$, for each of these conditions:

1. $\sum_{n=0}^\infty |c_{mn} | =\sum_{n=0}^\infty c_{mn} = 2^{-m}\sum_{n=0}^m \binom{m}{n} = 1$.
2. $c_{mn} = \frac{m!}{2^m n!(m-n)!} \le \frac{m^n}{n!2^m} \xrightarrow[m\to\infty]{} 0.$
3. see 1.

So the assumptions for Theorem 2 are verified, and the result holds.

(This type of result is a "regularity" result for a summation method, and such methods are the main focus of Hardy's book.)

Below is not entirely rigorous.

Since $\{a_n\}$ is convergent, we can pick $N$ such that $a_m$ is close to $z$ for any $m\ge N$.

Let $n$ large, and due to the fact that if $N=O(1)$ (implies $o(n)$), sum of $\sum_{k=1}^N {{n}\choose{k}}=O(n^N)$

We have $\frac{{n \choose 1} a_1 + {n \choose 2} a_2+ \cdots+ {n \choose n} a_n}{2^n} $ is close to $\frac{{n \choose 1} (z-\epsilon) + {n \choose 2} (z-\epsilon)+ \cdots+ {n \choose n} (z-\epsilon)}{2^n} $, close to $z$.

Let $\epsilon>0$ and define $b_n=a_n-z$. Since $a_n\to z$ there exists $N$ such that for all $n\geq N$, $|b_n|<\epsilon$. Now, consider the limit

$$\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}a_i=\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}(b_i+z)$$

$$=\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}b_i+\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}z=z+\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}b_i.$$

We may simplify the problem and only consider the remaining limit. Note that proving the original statement is equivalent to proving

$$\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}b_i=0.$$

For $n\geq N$ split the sum into two parts, $1\leq i<N$ and $i\geq N$. We have

$$\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}b_i=\lim_{n\to\infty}\left(\frac{1}{2^n}\sum_{i=1}^{N-1} \binom{n}{i}b_i+\frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}b_i\right)$$

$$=\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^{N-1} \binom{n}{i}b_i+\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}b_i.$$

Now, the first limit is clearly zero as $\binom{n}{i}=O(n^i)$ which implies the numerator is some polynomial in $n$ while the denominator is an exponential in $n$. Thus, our problem is distilled down to showing

$$\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}b_i=0.$$

To do this, note that

$$\frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}b_i\leq \frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}|b_i|<\frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}\epsilon\leq \frac{\epsilon}{2^n}\sum_{i=1}^{n} \binom{n}{i}=\epsilon.$$

Thus,

$$\frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}b_i$$

is bounded by every positive number and hence is $0$ in the limit. We conclude

$$\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}a_i=z.$$